If a polymer is prepared from 4.90008 g of acrylic acid, and 4.625 mL of 8.0 M NaOH, what will the mass of the "dry" polymer be after all of the water is removed? Assume all reactions go to completion. Hint: You may want to construct a BCA table to determine the outcome of the neutralization step.
The polymerization reaction is
The molar mass of acrylic acid is 72.0627 g/mol; the amount of substance is n = m/M = 4.90008/72.0627 = 0.068 moles.
The amount of NaOH is n = cV = 8.0*0.004625 = 0.037 moles. The sodium hydroxide is a limiting reactant.
(The mass of polymer) = (mass of NaOH) + (mass of acrylic acid) - (mass of water);
The mass of NaOH is n*M = 0.037*40 = 1.48 g;
The mass of acrylic acid is n*M = 0.037*72.0627 = 2.666 g;
The mass of water is n*M = 0.037*18.0153 = 0.6656 g;
The mass of polymer =1.48+2.666-0.6656 = 3.48 g.
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